Compute the volume of oxygen needed for complete combustio fo 114g of octane `(C_(8)H_(18))` at STP.

Explanation

To compute the volume of oxygen needed for the complete combustion of 114 g of octane (C₈H₁₈) at STP, we will follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of octane.
The balanced equation for the combustion of octane is:
\[ 2 \text{C}_8\text{H}_{18} + 25 \text{O}_2 \rightarrow 16 \text{CO}_2 + 18 \text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of octane.
To find the number of moles of octane, we use the formula:
\[ \text{Number of moles} (n) = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of octane (C₈H₁₈) can be calculated as follows:
- Carbon (C): 12 g/mol × 8 = 96 g/mol
- Hydrogen (H): 1 g/mol × 18 = 18 g/mol
- Total molar mass = 96 g/mol + 18 g/mol = 114 g/mol

Now, substituting the values:
\[ n = \frac{114 \text{ g}}{114 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the amount of oxygen required for the combustion of octane.
From the balanced equation, we see that 2 moles of octane require 25 moles of oxygen. Therefore, for 1 mole of octane, the oxygen required is:
\[ \text{Oxygen required} = \frac{25 \text{ moles O}_2}{2 \text{ moles C}_8\text{H}_{18}} \times 1 \text{ mole C}_8\text{H}_{18} = 12.5 \text{ moles O}_2 \]

### Step 4: Convert moles of oxygen to volume at STP.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of oxygen required is:
\[ \text{Volume of O}_2 = 12.5 \text{ moles O}_2 \times 22.4 \text{ L/mole} = 280 \text{ L O}_2 \]

### Final Answer:
The volume of oxygen needed for the complete combustion of 114 g of octane at STP is **280 liters**.

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